WBBSE Solutions For Class 10 Maths Chapter 5 Ration And Proportion Exercise 5.2

WBBSE Solutions For Class 10 Maths Chapter 5 Ration And Proportion Exercise 5.2

Question 1. Let us find the value of x for the following proportions :

1.10 : 35 :: x : 42

Solution: 10 : 35 :: x : 42

or, 10/35 = X/42

or, 35X = 10 x 42

X = 10 x 42/35

X = 420/35

X = 12

2. X : 50 :: 3 : 2

Solution: X : 50:: 3 : 2

or, 2X = 50 x 3

X= 50 x 3/2

X = 150/2

=75

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Question 2. Let us find the fourth proportional of the following:

1. 1/3, 1/4, 1/5

Solution: Given: 1/3, 1/4, 1/5

4th Proportional = \(\frac{\frac{1}{4} \times \frac{1}{5}}{1 / 3}\)

4th Proportional = 3/20

2. 9.6 kg, 7.6 kg, 28.8 kg.

Solution: Given: 9.6 kg, 7.6 kg, 28.8 kg.

4th proportional = 7.6kg x 28.8kg / 9.6kg

4th Proportional=22.8

3. \(x^2 y^2, y^2 z, z^2 x\)

Solution: Given: \(x^2 y^2, y^2 z, z^2 x\)

∴ 4th proportional = \(\frac{y^2 z \times z^2 x}{x^2 y}=\frac{y z^3}{x^2 y}\)

4. \((\mathbf{p}-\mathbf{q}),\left(\mathbf{p}^2-\mathbf{q}^2\right), \mathbf{p}^2-\mathbf{p q}+\mathbf{q}^2\)

Solution: \((\mathbf{p}-\mathbf{q}),\left(\mathbf{p}^2-\mathbf{q}^2\right), \mathbf{p}^2-\mathbf{p q}+\mathbf{q}^2\)

∴ 4 th proportional = \(\frac{\left(p^2-q^2\right)\left(p^2-p q+q^2\right)}{(p-q)}=(p+q)\left(p^2-p q+q^2\right)\)

= \(p^3+q^3\)

Question 3. Let us find the 3rd proportional of the following positive numbers

1. 5, 10

Solution: Given: 5, 10

3rd Proportional = \(\frac{10 \times 10}{5}=20 \text {. }\)

2. 0.24, 0.6

Solution: Given: 0.24, 0.6

3rd proportional = \(\frac{0.6 \times 0.6}{0.24}=\frac{36}{24}=\frac{3}{2}=1.5\)

3. \(\mathbf{p}^3 q^2, q^2 \mathbf{r}\)

Solution: \(\mathbf{p}^3 q^2, q^2 \mathbf{r}\)

3rd proportional = \(\frac{q^2 r \times q^2 r}{p^3 q^2}=\frac{q^4 r}{p^3 q^2}=\frac{q^2 r^2}{p^3}\)

4. \((x-y)^2,\left(x^2-y^2\right)^2\)

Solution: \((x-y)^2,\left(x^2-y^2\right)^2\)

∴ 3rd proportional = \(\frac{\left(x^2-y^2\right)^2\left(x^2-y^2\right)^2}{2}\)

= \(\frac{(x+y)^2(x-y)^2(x+y)^2(x-y)^2}{(x-y)^2}\)

= \((x+y)^2(x-y)^2\)

 

Question 4. Let us find the mean proportional of the following positive numbers

1. 5 and 80

Solution: 5 and 80

Required mean proportional = √5×80 = √400 = 20.

2. 8.1 and 2.5

Solution: 8.1 & 2.5

Mean proportional = √8.1×2.5 = √20.25 = 4.5.

3. x³y and xy³

Solution: x³y and xy³

Mean proportional √x³y.xy³

=√x⁴y⁴

=x²y²

4. (x − y)², (x + y)²

Solution: (x-y)², (x + y)²

Mean proportional = √(x-y)². (x+y)²

=(x-x) (x + y). 

Question 5. If the two ratios a: b and c : d express mutually opposite relations, let us write what relation will be expressed by their inverse relation.

Answer. Each other are inverse proportion.

Question 6. Let us write how many continued proportions can be constructed by three numbers in continued proportions.

Answer. Two.

Question 7. If the first and second of the five numbers in continued proportion are 2 and 6 respectively, let us find the fifth number.

Solution: Let 5 consecutive proportional numbers are a, b, c, d, e.

∴ \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{c}}=\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{d}}{\mathrm{e}}\)

Here, a = 2 & b = 6.

∴ \(c=\frac{6 \times 6}{2}=18\)

∴ \(\frac{6}{18}=\frac{18}{d}\)

∴ \(\mathrm{d}=\frac{18 \times 18}{6}=54\)

Again, \(\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{d}}{\mathrm{e}} \text { or, } \frac{18}{54}=\frac{54}{\mathrm{e}}\)

∴ \(e=\frac{54 \times 54}{18}=54 \times 3=162\)

 

Question 8. Let us write by calculating what should be added to each of 6, 15, 20, and 43 to make the sums proportional.

Solution: 6, 15, 20 & 43

Let x be the required number, which should be added to each term.

Then (6 + x) : (15 + x) :: (20 + x) : (43 + x)

or, \(\frac{6+x}{15+x}=\frac{20+x}{43+x}\)

or, (6 + x)(43 + x) = (15 + x)(20 + x)

or, \(258+43 x+6 x+x^2=300+20 x+15 x+x^2\)

or, 49x – 35x = 300 – 258

or, 14x = 42

∴ \(x=\frac{42}{14}=3\)

∴ The required number = 3.

Question 9. Let us find what should be subtracted from each of 23, 30, 57, and 78 to make the results proportional.

Solution: Let x be the number which should be subtracted from each of the terms & then they will be proportional.

∴ (23 – x) : (30 – x) :: (57 – x) : (78 – x)

or, \(\frac{23-x}{30-x}=\frac{57-x}{78-x}\)

or, (30 – x)(57 – x) = (23 – x)(78 – x)

or, \(1710-57 x-30 x+x^2=1794-78 x-23 x+x^2\)

or, 87x + 101x = 1794 – 1710

or, 14x = 84

∴ \(x=\frac{84}{14}=6\)

∴ The required number = 6.

 

Question 10. Let us find what should be subtracted from each of p, q, r, and s to have four numbers in proportion.

Solution: p, q, r, s.

Let x be the number which should be subtracted from each of the terms so that they will be proportional.

∴ (p – x) : (q – x) : (r – x) : (s – x)

or, \(\frac{p-x}{q-x}=\frac{r-x}{s-r}\)

or, (q – x)(r – x) = (p – x)(s – x)

or, \(q r-r x-q x+x^2=p s-p s-s x+x^2\)

or, px + sx – rx – qx = ps – qr

or, x(p + s – r – q) = ps – qr

∴ \(x=\frac{p s-q r}{p+s-r-q}\)

∴ The required number = \(\frac{p s-q r}{p+s-r-q}\)

 

WBBSE Solutions For Class 10 Maths Chapter 5 Ration And Proportion Exercise 5.2 Applications

Question 1. If 5: 4: 10: 8, then with the help of componend and dividend of proportion, we get, (5+ 4): (5-4)::(10+8):

Solution: (5+ 4) : (5-4) = (10+8): (10-8)

Question 2. If a b::c: d, prove that (4a + 7b): (4a7b) :: (4c+ 7d): (4c-7d)

Solution: If a : b :: c : d,

Prove that (4a + 7b): (4a7b) :: (4c+ 7d): (4c-7d)

Let \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{d}}=\mathrm{k}\)       [k ≠ 0]

∴ a = bk and c = dk

L.H.S = \(\frac{4 a+7 b}{4 a-7 b}=\frac{4 \times b k+7 b}{4 \times d k-7 b}=\frac{4 b k+7 b}{4 d k-7 b}=\frac{b(4 k+7)}{b(4 k-7)}\)

= \(\frac{4 \mathrm{k}+7}{4 \mathrm{k}-7}\)

R.H.S = \(\frac{4 \mathrm{c}+7 \mathrm{~d}}{4 \mathrm{c}-7 \mathrm{~d}}=\frac{4 \times \mathrm{dk}+7 \mathrm{~d}}{4 \times \mathrm{dk}-7 \mathrm{~d}}=\frac{\mathrm{d}(4 \mathrm{k}+7)}{\mathrm{d}(4 \mathrm{k}-7)}\)

= \(\frac{4 \mathrm{k}+7}{4 \mathrm{k}-7}\)

∴ L.H.S = R.H.S

 

Question 3. We can write by applying the addenda property of proportion, 2/3 = 6/9 = 8/12 = 2+6+8 / + + .

Solution: 2/3 = 6/9 = 8/12

= 2+6+8/ 3+9+12

= 16/24

Question 4. \(\frac{x}{a+b-c}\) \(\frac{\mathrm{y}}{\mathrm{b}+\mathrm{c}-\mathrm{a}}\) \(\frac{z}{c+a-b}\), let us prove that each ratio = \(\frac{x+y+z}{a+b+c}\)

Solution: \(\frac{x}{a+b-c}\)

= \(\frac{\mathrm{y}}{\mathrm{b}+\mathrm{c}-\mathrm{a}}\)

= \(\frac{z}{c+a-b}\)

By applying addendo,

Each ratio = \(\frac{x+y+z}{a+b-c+b+c-a+c+a-b}\)

= \(\frac{x+y+z}{a+b+c}\)

Question 5. If (4a+5b) (4c-5d) = (4a-5b) = (4a-5b) (4c + 5d), let us prove that a, b, c, and d are in proportion.

Solution:

Given

If (4a+5b) (4c-5d) = (4a-5b) = (4a-5b) (4c + 5d

(4a+5b) (4c-5d) = (4a-5b) = (4a-5b) (4c + 5d)

or, 16ac +20bc20ad25bd = 16ac – 20bc + 20ad – 25bd

or, 20bc+20bc = 20ad + 20ad

or, 40bc = 40ad

or, bc = ad

a/b = c/d

a, b, c & d are in proportion